Question: You have found the following ages (in years) of all 6 zebras at your local zoo: $ 2,\enspace 24,\enspace 15,\enspace 13,\enspace 5,\enspace 19$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{2 + 24 + 15 + 13 + 5 + 19}{{6}} = {13\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-11$ years $121$ years $^2$ $24$ years $11$ years $121$ years $^2$ $15$ years $2$ years $4$ years $^2$ $13$ years $0$ years $0$ years $^2$ $5$ years $-8$ years $64$ years $^2$ $19$ years $6$ years $36$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{121} + {121} + {4} + {0} + {64} + {36}} {{6}} $ $ {\sigma^2} = \dfrac{{346}}{{6}} = {57.67\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{57.67\text{ years}^2}} = {7.6\text{ years}} $ The average zebra at the zoo is 13 years old. There is a standard deviation of 7.6 years.